Probability Questions

course Mth 152

Please see questions below. I will be taking my two remaining tests on Monday or Tuesday. Between my work schedule and the availability of the proctor, I have been having difficulty making arrangements to take the tests, but my work supervisor has promised that I can leave early on Monday and/or Tuesday and the proctor has informed me that he will be at the college.

1) I know I can find the number of possible poker hands by C(52,5), but how do I find the number of possible

'four of a kind' or 'full house' hands?

To get four of a kind in a given denomination you have to get all four of that denomination, then one of the remaining cards. There would be C(4, 4) * C(48, 1) ways to do this.

There are 13 possible denominations, so there are 13 * C(4, 4) * C(48, 1) ways to get four of a king.

A full house consists of 3 of one denomination and 2 of another. There are C(4, 3) ways to get 3 cards of a given denomination, and C(4, 2) ways to get 2 cards of another. So for two given denominations there are C(4, 3) * C(4, 2) ways to get the corresponding full house. There are P(13, 2) choices for the two denominations (13 possibilities for the first denomination, the one of which you get 3 cards, and 12 for the second) so there are P(13, 2) * C(4, 3) * C(4, 2) ways to get a full house.

2) In a population, 30% of females are pure dominant, 60% are hybrid, and 10% are pure recessive. If a pure recessive male mates with a random female and their first offspring has the dominant trait, what is the probability that the female is pure dominant?

Given that the offspring has the dominant trait, we can say the following:

The dominant trait had to come from the female, since the male was pure recessive.

The female was therefore either hybrid or pure dominant.

If the female was hybrid she would have a 50% chance of passing on the gene.

If the female was pure dominant she would have a 100% chance of passing on the gene.

There are twice as many hybrid females as pure dominant females.

So the probability the female was hybrid is equal to the probability that the female was pure dominant.